subspace structure of some operator and banach...

Journal of Logic & Analysis 3:2 (2011) 1–31ISSN 1759-9008

1

Subspace structure of some operator and Banach spacesTIMUR OIKHBERG

CHRISTIAN ROSENDAL

Abstract: We construct a family of separable Hilbertian operator spaces, such thatthe relation of complete isomorphism between the subspaces of each member ofthis family is complete Kσ . We also investigate some interesting properties ofcompletely unconditional bases of the spaces from this family. In the Banachspace setting, we construct a space for which the relation of isometry of subspacesis equivalent to equality of real numbers.

2000 Mathematics Subject Classification 46L07 (primary); 03E15 (secondary)

Keywords: operator space, isomorphism between subspaces, Effros-Borel struc-ture, classification

1 Introduction and main results

Recently, there has been much progress in describing the complexity of various relationsbetween subspaces of a given separable Banach space. The reader is referred to [4, 5, 6]for the known results on the relations of isomorphism, biembeddability, and more.Isometry and local equivalence (finite representability) are handled in [16] and [7],respectively.

In this paper, we consider an operator space analogue of this problem (see Section 2 fora brief introduction into operator spaces). The Effros-Borel structure on the set S(Z) ofinfinite-dimensional subspaces of a separable operator space Z is defined in the sameway as for Banach spaces, see e.g. [11, Chapter 12], or [6]. Reasoning as in Section 2of [6], we show that the relations of complete isomorphism, complete biembeddability,and such, defined on S(Z), are analytic equivalence relations.

In the commutative space setting, the famous Gowers-Komorowski-Tomczak Theorem(see e.g. [25]) shows that any separable Banach space, isomorphic to all of its infinitedimensional subspaces, must be isomorphic to �2 . The space �2 can be equipped withuncountably many 1-homogeneous operator space structures (an operator space X iscalled 1-homogeneous if the equality �u� = �u�cb holds for any u ∈ B(X)). For

Published: March 2011 doi: 10.4115/jla.2011.3.2

http://www.ams.org/mathscinet/search/mscdoc.html?code=46L07,(03E15)

http://dx.doi.org/10.4115/jla.2011.3.2

2 Timur Oikhberg and Christian Rosendal

instance, the row and column spaces R and C, as well as their complex interpolationspaces (R,C)θ , have this property. These, and other, examples of homogeneousHilbertian spaces can be found, for instance, in [21, 22] (an operator space is calledHilbertian if it is isomorphic to a Hilbert space). Clearly, any 1-homogeneous operatorspace, having �2 as its underlying Banach space, is completely isometric to all of itsinfinite dimensional subspaces. For such spaces, the relation of complete isomorphismbetween infinite dimensional subspaces is trivial.

It is not known how “simple” the relation of complete isomorphism between subspacesof X may get if X is not homogeneous (or not isomorphic to a Hilbert space). Ourmain result provides a partial answer to this question.

Theorem 1.1 There exists a family F of operator spaces, such that, for any operatorspace X ∈ F, the following is true:

(1) X is isometric to �2 .(2) The relation of complete isomorphism and complete biembeddability on S(X)

are Borel bireducible to the complete Kσ relation.

The family F contains a continuum of operator spaces, not completely isomorphic toeach other.

It is possible to prove that the relation of complete isometry on S(X) (X ∈ F) isBorel bireducible to the equality on R. The proof proceeds along the same lines asTheorem 6.1, but is exceedingly technical, and not very illuminating. We thereforeomit it, and present a related Theorem 1.4 instead.

Each space from F has its canonical basis (defined in Section 4), which is 1-completelyunconditional. It turns out these bases have interesting properties of their own.

Theorem 1.2 Any subspace Y of an operator space X from the family F has 1-completely unconditional canonical basis. Any C-completely unconditional basis insuch a Y is φ(C)-equivalent (up to a permutation) to the canonical basis of Y , withφ(C) polynomial in C .

For certain spaces X , a stronger result holds.

Theorem 1.3 For any a > 1 there exists an operator space X , belonging to the familyF, such that the canonical basis in any subspace of X is a-equivalent to a subsequenceof the canonical basis of X . Consequently, every C-completely unconditional basicsequence in X is φ(C)-equivalent (up to a permutation) to a subsequence of thecanonical basis, with φ(C) polynomial in C .

Journal of Logic & Analysis 3:2 (2011)

Subspace structure of some operator and Banach spaces 3

Note that homogeneous Hilbertian operator spaces satisfy the conditions of the threetheorems above. The spaces from the family F are not homogeneous, yet, by Theorems1.2 and 1.3, they share certain essential properties of homogeneous spaces.

As part of our motivation lies in the field of Banach spaces, we mention a few classicalresults related to Theorems 1.1, 1.2, and 1.3.

No “commutative” counterpart of Theorem 1.1 has yet been obtained: for a Banachspace E , very little is known about the “upper” estimates on the complexity of theisomorphism relation on S(E). It is possible that the relation of isomorphism on S(E)is complete analytic whenever E is a separable Banach space, not isomorphic to �2 .

Theorem 1.2 shows (among other things) that every subspace of X has an unconditionalbasis. In the commutative setting, �2 is the known only separable Banach space withthe property that each of its separable subspaces has an unconditional basis. It is along-standing open question (see e.g. [8, p. 203]) whether there exist other spacessharing the same property. If a separable E has this property, it must be “close to” �2 .First, any subspace of E has the Approximation Property, therefore, by [15, Theorem1.g.6], E has to have type 2 − ε and cotype 2 + ε for any ε > 0. Furthermore, by[20, Theorem 10.13], E has weak cotype 2. By [12], E is �2 -saturated. Finally, ifE = �2(F) for some Banach space F , then E is isomorphic to a Hilbert space [13].

Searching for a “commutative” analogue of Theorem 1.3, we strengthen the questionposed in the previous paragraph: suppose a Banach space E has an unconditionalbasis (ei), such that every subspace of E has an unconditional basis, equivalent to asubsequence of (ei). Must E be isomorphic to �2 ?

Finally, in the Banach space setting, we construct a space E , not isometric to �2 , butwhose subspace structure is “very simple.”

Theorem 1.4 The following equivalence relations between infinite dimensional sub-spaces of R⊕1 �2 (or, in the complex case, C⊕1 �2 ) are Borel bireducible to (R,=):(i) isometry, (ii) having Banach-Mazur distance 1; (iii) isometric bi-embeddability;(iv) almost isometric bi-embeddability.

Recall that the Banach-Mazur distance between Y and Z is defined as d(Y, Z) =inf{�u��u−1� : u ∈ B(Y, Z)}. Y is called almost isometrically embeddable into Zif for every ε > 0 there exists Yε �→ Z s.t. d(Y, Yε) < 1 + ε (d(·, ·) denotes theBanach-Mazur distance between spaces).

The rest of the paper is organized as follows: Section 2 provides a brief introductioninto operator spaces and c.b. maps. In Section 3 we construct, for each contraction



A ∈ B(�2), a subspace X(A) of R ⊕ C, and study the properties of the spaces X(A).Further investigation is carried out in Section 4, where we concentrate on the case whenA is compact. Unconditional bases in the spaces X(A) are described in Section 5. InSection 6 we show that, for the “right” compact contractions A, the relations of completeisomorphism and complete biembeddability on S(X(A)) are Borel bireducible to thecomplete Kσ relation. Having gathered all the preliminary results, we prove Theorems1.1, 1.2, and 1.3 in Section 7. Finally, in Section 8 we prove Theorem 1.4.

1.1 Acknowledgements

The first author was partially supported by CORCL grant from the UC system. Thesecond author was partially supported by NSF grants DMS 0901405 and DMS 0919700.We would like to thank the referee for valuable comments, and, in particular, forbringing the reference [8] to our attention.

2 Introduction into operator spaces

As our paper deals primarily with operator spaces, we are compelled to recall somebasic definitions and facts about the topic. An interested reader is referred to [3, 19, 22]for more information. A (concrete) operator space X is, for us, just a closed subspaceof B(H) (H is a Hilbert space). If X and Y are operator spaces, embedded into B(H)and B(K) respectively, we defined the minimal tensor product of X and Y (denotedsimply by X ⊗ Y ) as the closure of the algebraic tensor product X ⊙ Y in B(H ⊗2 K).It is common to denote Mn ⊗ X by Mn(X). Here, Mn = B(�n

2) is the space ofn × n matrices. We view Mn(X) as the space of X -valued n × n matrices, with thenorm � · �n . It is easy to see that the sequence of matricial norms � · �n satisfiestwo properties (Ruan’s axioms): (i) for any v ∈ Mn(X), α ∈ Mn,k , and β ∈ Mk,n ,�(β ⊗ IX)v(α ⊗ IX)�k � �β��x�n�α�, and (ii) for any v ∈ Mn(X) and w ∈ Mk(X)�,�v ⊕ w�k+n = max{�v�n, �w�k}. It turns out that the converse is also true. SupposeX is a matricially normed space – that is, it is a Banach space, for which the spacesMn(X) of n × n X -valued matrices are equipped with the norms � · �n , satisfying (i)and (ii) above. Then the norms � · �n arise from an isometric embedding of X intoB(H), for some Hilbert space H . The spaces X as above are sometimes called abstractoperator spaces.

It is easy to see that a subspace of an operator space is, again, an operator space (we usethe notation �→ to denote one operator space being a subspace of another). Moreover,



a quotient and the dual of an operator space can be again equipped with an operatorspace structure. Once again, the reader is referred to [3, 19, 22] for details.

A map u from an operator space X to an operator space Y is called completely bounded(c.b. for short) if its c.b. norm

�u�cb = supn

�IMn ⊗ u�B(Mn(X),Mn(Y)) = �IB(�2) ⊗ u�B(B(�2)⊗X,B(�2)⊗Y)

is finite. The set of all c.b. maps from X to Y is denoted by CB(X, Y). Clearly,�u�cb ≥ �u�, and CB(X, Y) ⊂ B(X, Y) (the inclusion may be strict). The operatorspaces X and Y are called completely isomorphic (completely isometric) if there existsu ∈ CB(X, Y) such that u−1 is c.b. (resp. �u�cb = �u−1�cb = 1). We shall usethe notation � for complete isomorphism. Moreover, we say that X is c-completelyisomorphic to Y (X

c� Y ) if there exists u ∈ CB(X, Y) with �u�cb�u−1�cb � c.

Complete embeddability and biembeddability are defined in the obvious way.

Suppose the operator space s X and Y are embedded into B(H) and B(K), respectively.We define the direct sum of operator spaces X and Y (denoted by X ⊕∞ Y , orsimply X ⊕ Y ) by viewing X ⊕ Y as embedded into B(H ⊕2 K). Note that anyu ∈ Mn(X ⊕ Y) has a unique expansion as v ⊕ w, with v ∈ Mn(X) and w ∈ Mn(Y).Then �u� = max{�v�, �w�}.

Throughout this paper, we work with the row and column spaces. Recall that a Hilbertspace H can be equipped with row and column operator space structure, denoted byHR and HC , respectively. The space HR is defined as the linear space of operatorsξ⊗ ξ0 , where ξ0 is a fixed unit vector, and ξ runs over H . Here, for ξ ∈ H and η ∈ K ,ξ ⊗ η denotes the operator in B(H,K), defined by (ξ ⊗ η)ζ = �ζ, ξ�η . Similarly, thespace HC is defined as the space of operators ξ0 ⊗ ξ (ξ ∈ H ).

It is easy to see that, if K is a subspace of H , then KC (KR ) is a subspace of HC(resp. HR ). For simplicity of notation, we write denoted by R and C, instead of (�2)Rand (�2)C , respectively. One can use matrix units to describe these spaces. We denoteby Eij ∈ B(�2) the infinite matrix with 1 on the intersection of the i-th row and thej-th column, and zeroes elsewhere. Then R (C) is the closed linear span of (E1j)∞j=1(respectively, (Ei1)∞i=1)). Below we list a few useful properties of row and columnspaces. Here, H and K are Hilbert spaces.

(1) HR and HC are isometric to H (as Banach spaces).

(2) For any u ∈ B(H,K), �u� = �u�CB(HR,KR) = �u�CB(HC,KC) .



(3) If (ξi) is an orthonormal system in H , then, for any finite sequence (ai) ofelements of Mn ,

��

i ai ⊗ ξi�Mn(HR) = �(�

i aia∗i )1/2� = ��

i aia∗i �1/2,��

i aj ⊗ ξi�Mn(HC) = �(�

i a∗i ai)1/2� = ��

i a∗i ai�1/2.

(4) Suppose H(1), . . . ,H(n) are Hilbert spaces. Then the formal identity map id :(H(1) ⊕2 . . . ⊕2 H(n))R → H(1)

R ⊕ . . . ⊕ H(n)R is a complete contraction, and

�id−1�cb � √n. The same is also true for column spaces.

(5) For any u ∈ CB(HR,KC) or u ∈ CB(HC,KR), �u�2 = �u�cb (here, � · �2 is theHilbert-Schmidt norm).

(6) Duality: H∗R = HC , and H∗

C = HR .

(7) For any operator space X and any u ∈ B(X,HC) (u ∈ B(X,HR)), �u�cb =�IC ⊗u�B(C⊗X,C⊗HC) (resp. �u�cb = �IR ⊗u�B(R⊗X,R⊗HR) ). This result followsfrom the proof of Smith’s lemma – see e.g. Proposition 2.2.2 of [3].

(8) If H is separable infinite dimensional, then HR (HC ) is completely isometric toR (resp. C).

A sequence (xi)i∈I in an operator space X is called normalized if �xi� = 1 for everyi ∈ I . (xi) is said to be a c-completely unconditional basic sequence (c ≥ 1) if, for anyfinite sequence of matrices (ai), and any sequence of scalars λi ∈ {λ ∈ C : |λ| � 1},we have �

�i λiai ⊗ xi�Mn(X) � c�

�i ai ⊗ xi�Mn(X) . It is easy to see that any such

sequence (xi) is linearly independent. A c-completely unconditional basic sequence(xi) ⊂ X is called a c-completely unconditional basis in X if X = span[xi : i ∈ I](here and below, span[F] refers to the closed linear span of the family F ).

A convenient example of a normalized 1-completely unconditional basis is provided byan orthonormal basis in R or C. For future reference, we observe that any completelyunconditional basis is “similar to” an orthogonal one. More precisely, suppose X isan operator space, which is isometric to a Hilbert space (this is the setting we areconcerned with in this paper). Suppose, furthermore, that (xi) is a normalized c-completely unconditional basis in an operator space X , isometric to a Hilbert space,then (xi) is “similar to” an orthonormal basis: for any finite sequence of scalars (αi),

c2�

�

i

αixi�2≥ Ave±�

�

i

±αixi�2 =

�

i

|αi|2,

and similarly, c−2��

i αixi�2 � �

i |αi|2 . Thus, there exists an U : X → �2(I), s.t.

(Uxi)i∈I is an orthonormal basis, and �U�, �U−1� � c.

Families (xi)i∈I and (yi)i∈I in operator spaces X and Y , respectively, are called c-equivalent if there exists a map u : span[xi : i ∈ I] → span[yi : i ∈ I], such



that uxi = yi , and �u�cb�u−1�cb � c. Two sequences are equivalent if they are c-equivalent, for some c. Furthermore, (xi)i∈I is c-equivalent to a subfamily of (yj)j∈J ifthere exists a subset I� ⊂ J , such that |I| = |I�|, and (xi)i∈I is c-equivalent to (yj)j∈I� .Finally, (xi)i∈I is c-equivalent to (yi)i∈I up to a permutation if there exists a bijectionπ : I → I such that (xi)i∈I is c-equivalent to (yπ(i))i∈I .

A sequence (xi)i∈I is called completely unconditional if it is c-completely uncondi-tional, for some c. A completely unconditional basis, equivalence of sequences etc.are defined by dropping the c, in a similar manner.

3 Subspaces of R ⊕ C: basic facts

Suppose H and K are separable Hilbert spaces, and A ∈ B(H,K) is a contraction.Denote by XR(H,K,A) the subspace of HR ⊕KC , spanned by (e,Ae) (e ∈ H ). If thereis no confusion as to the spaces H and K , we simply write XR(A). XC(A) is definedin the same way. Often, we write X(A) (X(H,K,A)) instead of XR(A) (XR(H,K,A)).Note that the formal identity map id : H → X(H,K,A) : ξ �→ ξ ⊕ Aξ is an isometry.Thus, we identify subspaces of X(H,K,A) with those of H . This identification givesmeaning to the notation A|Y , where Y �→ X(A).

Remark 3.1 Although the spaces R and C are “simple”, the structure of their directsum R ⊕ C is rather rich. For instance, it was shown in [27] that the “operator Hilbertspace” OH is a subspace of a quotient of R ⊕ C (actually, the results of that paper aremuch more general). It follows from [27] that the spaces X(A) (A ∈ B(�2)) defined asabove are natural “building blocks” of subspaces of R ⊕ C.

Begin by stating a simple lemma.

Lemma 3.2 Suppose H , K , and K� are Hilbert spaces, and A ∈ B(H,K) is a contrac-tion.

(1) Suppose U ∈ B(K,K�) is such that �Uξ� = �ξ� for any ξ ∈ ran A. ThenX(H,K,A) is completely isometric to X(H,K�,UA). In particular, X(H,K,A) iscompletely isometric to X(H,H, |A|), where |A| = (A∗A)1/2 .

(2) Suppose P1, . . . ,Pm and Q1, . . . ,Qm are families of orthogonal mutually orthog-onal projections on H and K , respectively, such that

�k Pk = IH ,

�k Qk = IK ,

and A =�m

k=1 QkAPk . Set Hk = Pk(H) and Kk = Qk(K). Then theformal identity operator id from X(H,K,A) to X(H1,K1,Q1AP1) ⊕ . . . ⊕X(Hm,Km,QmAPm) is a complete contraction, and �id−1�cb � √

m.



(3) X(A) is |λ|−1 -completely isomorphic to X(λA) whenever 0 < |λ| � 1.

Proof We only establish part (2). A Gram-Schmidt orthogonalization shows that anyelement x ∈ Mn(H) can be written as x = x1 + . . . + xn , with xk =

�i aki ⊗ ξki

aki ∈ Mn , and (ξki) a finite orthonormal systems in Hk . Then

�x�Mn(X(A)) = max {��

k

�

i

aki ⊗ ξki�Mn(HR), ��

k

�

i

aki ⊗ Aξki�Mn(KC)}.

By the basic properties of row and column spaces (listed at the end of the previoussection),

maxk

�

�

i

aki ⊗ ξki�Mn(HR) � �

�

k,i

aki ⊗ ξki�Mn(HR)

�√

m maxk

�

�

i

aki ⊗ ξki�Mn(HR).

Furthermore, the vectors Aξki belong to the mutually orthogonal spaces Kk , hence

maxk

�

�

i

aki ⊗ Aξki�Mn(HC) � �

�

k,i

aki ⊗ Aξki�Mn(HC)

�√

m maxk

�

�

i

aki ⊗ Aξki�Mn(HC).

Note that

�xk�Mn(X(Hk,Qk,QkAPk))

= max {�

�

i

aki ⊗ ξki�Mn(HR), ��

i

aki ⊗ Aξki�Mn(HC)}.

Therefore,

maxk

�xk�Mn(X(Hk,Qk,QkAPk))

= max{maxk

�

�

i

aki ⊗ ξki�Mn(HR),maxk

�

�

i

aki ⊗ Aξki�Mn(HC)}

� �x�Mn(X(A))

�√

m max{maxk

�

�

i

aki ⊗ ξki�Mn(HR),maxk

�

�

i

aki ⊗ Aξki�Mn(HC)}.

We complete the proof by noting that

�x�Mn(X(H1,K1,Q1AP1)⊕...⊕X(Hm,Km,QmAPm)) = max1�k�m

�xk�Mn(X(Hk,Qk,QkAPk)).

Next, we establish our key tool for computing c.b. norms.



Lemma 3.3 Suppose A and B are contractions, and T ∈ B(X(A),X(B)). Then

�T�cb = max {�T�, sup{�BTu�2 : u ∈ B(�2,X(A)), �Au�2 � 1, �u� � 1}}.

Proof By definition, �T�cb = max{�T�CB(X(A),R), �BT�CB(X(A),C)}. To estimate thefirst term, note that id : X(A) → R is a complete contraction, hence

�T�CB(X(A),R) = �T ◦ id�CB(X(A),R) � �T�CB(R,R)�id�CB(X(A),R) = �T�.

However, �T� � �T�CB(X(A),R) , hence �T� = �T�CB(X(A),R) .

Next we estimate �BT�CB(X(A),C) . We know that

�BT�CB(X(A),C) = sup {�(IC ⊗ BT)(x)�C⊗C : x ∈ C ⊗ X(A), �x�C⊗X(A) � 1}.

Identifying elements of C ⊗ X(A) and C ⊗ C with operators from R to X(A) and C,respectively, we see that

�BT�CB(X(A),C) = sup {�BTu�2 : u ∈ CB(R,X(A)), �u�cb � 1}.

But �u�cb = max{�u�CB(R), �Au�CB(R,C)} = max{�u�, �Au�2}.

Next we examine the exactness of X(A)∗ . Recall that an operator space X is called exactif there exists c > 0 such that, for any finite dimensional subspace E of X , there existsan operator u from E to a subspace F of Mn (n ∈ N), such that �u�cb�u−1�cb < c.The infimum of all the c’s like this is called the exactness constant of X . It is easy tosee that R ⊕ C is 1-exact, hence so is X(A). The case of the dual is different.

Proposition 3.4 Suppose A ∈ B(H,K) is a contraction. Then the exactness constantof X(A)∗ is at least 2−5/2�A�2 . In particular, X(A)∗ is not exact if A is not Hilbert-Schmidt.

Proof Let c = 25/2ex(X(A)∗). By Corollary 0.7 of [23], there exist operators T1 :X(A) → R, T2 : R → X(A), S1 : X(A) → C, and T2 : C → X(A), such thatid = S2S1 + T2T1 (id is the identity on X(A)), and

�S1�cb�S2�cb + �T1�cb�T2�cb � max{�S1�cb, �T1�cb}(�S2�cb + �T2�cb) � c

By Lemma 3.3, �T2�cb ≥ �AT2�2 , and a simple calculation yields

�S2�cb ≥ �S2�CB(C,R) = �S2�2 ≥ �AS2�2.

Therefore,

�S1�cb�S2�cb + �T1�cb�T2�cb ≥ �S1��AS2�2 + �T1��AT2�2

≥ �A(S2S1 + T2T1)�2 = �A�2,

which implies the desired estimate for c.



Corollary 3.5 The space X(A) is completely isomorphic to R if and only if A isHilbert-Schmidt. Furthermore, the formal identity map id : R → X(A) is completelycontractive, and �id−1�cb = max{1, �A�2}.

Proof The estimates on the c.b. norm of id and id−1 follow from Lemma 3.3. Thus,

X(A)max{1,�A�2}

� R whenever A is Hilbert-Schmidt. On the other hand, if A is notHilbert-Schmidt, then, by Proposition 3.4, X(A)∗ is not exact. However, R∗ = C isexact, thus X(A) is not completely isomorphic to R.

We say that A ∈ B(H,K) (H and K are Hilbert spaces) is diagonalizable if theeigenvectors of |A| = (A∗A)1/2 span H . Equivalently, there exist orthonormal systems(ξi) and (ηi) in H and K , respectively, and a sequence (λi) of non-negative numbers,such that A =

�i λiξi ⊗ ηi .

When working with X(A), it is often convenient to have A diagonalizable. While everycompact operator is diagonalizable, a non-compact one need not have this property.However, we have:

Lemma 3.6 Suppose H and K are separable Hilbert spaces, A ∈ B(H,K) is acontraction, and ε > 0. Then there exists a diagonalizable contraction B ∈ B(H,K)such that �A − B�2 � ε, and X(A) is (1 + ε)2 -completely isomorphic to X(B). If A isnon-negative, B can be selected to be non-negative, too.

Proof By Lemma 3.2(1), it suffices to consider the case of A = |A|, and H = K . By[26], there exists a selfadjoint diagonalizable C ∈ B(H) such that �A − C�2 < ε/2.Let (ci) be the eigenvalues of C , and (ξi) the corresponding norm 1 eigenvectors.Define the operator B by setting Bξi = biξi , where

bi =

ci 0 � ci � 10 ci < 01 ci > 1

.

We claim that �B − C�2 < ε/2. Indeed, let I = {i : bi �= ci}. Then

�B − C�22 =

�

i∈I|bi − ci|

2 =�

i:ci<0

|ci|2 +

�

i:ci>1

|bi|2.

But �ξi,Aξi� ∈ [0, 1], hence, for ci < 0, |�ξi,Aξi� − �ξi,Cξi�| ≥ |ci|. Similarly, forci > 1, |�ξi,Aξi� − �ξi,Cξi�| ≥ |1 − ci|. Thus,

ε2

4> �A − C�

22 =

�

i,j

|�ξi, (A − C)ξj�|2≥

�

i

|�ξi, (A − C)ξi�|2≥

�

i∈I|bi − ci|

2.



By the triangle inequality, �A − B�2 < ε. Moreover, B is a non-negative contraction.

Denote the the formal identity map from X(A) to X(B) by U . It remains to show that�U�cb, �U−1�cb � 1 + ε. As U is an isometry, Lemma 3.3 implies

(3–1) �U�cb = sup{�Bu�2 : u ∈ B(�2,X(A)), �Au�2 � 1, �u� � 1}.

By the triangle inequality,

�Bu�2 � �Au�2 + �(B − A)u�2 � �Au�2 + �B − A�2 � 1 + ε,

hence (3–1) yields �U�cb � 1 + ε. �U−1�cb is estimated similarly.

Proposition 3.7 Suppose H and K are separable Hilbert spaces, B ∈ B(H,K) is acontraction, and 0 ∈ σess(|B|). Then X(B) is 4

√2-completely isomorphic to X(B)⊕R.

Proof We can assume B = |B|. By Lemma 3.6, there exists a non-negative diag-

onalizable contraction A such that A − B is Hilbert-Schmidt, and X(A)21/4

� X(B).As the essential spectrum is stable under compact perturbations, 0 ∈ σess(A). WriteA = diag (α), where α = (αi)i∈I , and diag (α) ∈ B(�2) is the diagonal operatorwith α,α2, . . . on the main diagonal. Then 0 is a cluster point of the set (αi). De-note the norm 1 eigenvectors, corresponding to αi , by ξi . Find I0 ⊂ I such that�

i∈I0α2

i < 1. Let I1 = I\I0 , and let P0 and P1 be the orthogonal projections ontoH0 = span[ξi : i ∈ I0] and H1 = span[ξi : i ∈ I1], respectively. By Lemma 3.2,X(A) is

√2-completely isomorphic to X(H0,H0,P0AP0) ⊕ X(H1,H1,P1AP1). By

Corollary 3.5, X(H0,H0,P0AP0) is completely isometric to R. Thus,

X(A)√

2� X(H0,H0,P0AP0) ⊕ X(H1,H1,P1AP1)

√2

� R ⊕ X(H1,H1,P1AP1)√

2� R ⊕ R ⊕ X(H1,H1,P1AP1)

= R ⊕ X(H0,H0,P0AP0) ⊕ X(H1,H1,P1AP1)√

2� R ⊕ X(A).

To summarize, X(A)4� X(A) ⊕ R. As X(A)

21/4

� X(B), we are done.

Corollary 3.8 Suppose H and K are separable infinite dimensional Hilbert spaces,and a contraction A ∈ B(H,K) satisfies 0 ∈ σess(|A|). Consider the operator A = A⊕0

from H = H ⊕2 �2 to K = K ⊕2 �2 . Then X(A)8� X(A).

Proof Let P be the orthogonal projection from H onto H . onto H . By Lemma 3.2(2)

and Corollary 3.5, X(A)√

2� X(A) ⊕ R. However, by Proposition 3.7, X(A)

4√

2�

X(A) ⊕ R.



4 Classification of subspaces using sequences

In this section, we study the spaces X(A) when A is compact, and establish a connectionbetween such spaces and a certain family of sequences. We denote by C the space ofcompact contractions A ∈ B(�2), which are not Hilbert-Schmidt. We denote by F theset of all spaces X(A), with A ∈ C.

Start by defining the canonical basis in X(A), where A ∈ B(H,K) is a compactcontraction. As X(A) = X(|A|), we assume henceforth that A = |A|, and H = K . Let(ξi)i∈I1 be the normalized eigenvectors of A, corresponding to the positive eigenvaluesof A. Furthermore, set H� = ker A, and let (ξi)i∈I0 be an orthonormal basis in H� (weassume that I1∩ I0 = ∅). Let I = I0∪ I1 . The vectors ei = ξi⊕Aξi ∈ HR⊕HC (i ∈ I )span X(A). Moreover, �ei� = 1 for each i, and, for any finite sequence (ai) ⊂ Mn ,

(4–1) �

�

i

ai ⊗ ei�2Mn(X(A)) = max

��

�

i

aia∗i �, ��

i

�Aξi�2a∗i ai�

�.

We say that the vectors ei = ei[A] form the canonical basis of X(A).

Now suppose (αi)i∈N is a sequence of numbers in [0, 1]. In an effort to link oper-ator spaces with certain sequences of scalars, we define the operator space Xd(α) =X(diag (α)). To describe the operator space structure of Xd(α), denote by (ξi) thecanonical orthonormal basis of �2 . Then the vectors ei(α) = ei = ξi ⊕αiξi ∈ HR ⊕KCform a 1-completely unconditional orthonormal basis in Xd(α), with

(4–2) �

�

i

ai ⊗ ei�2 = max{�

�

i

aia∗i �, ��

i

α2i a∗i ai�}

for any finite sequence of matrices (ai). We call (ei(α))i∈N the canonical basis ofXd(α). The formal identity from Xd(α) to Xd(β) is the linear operator mapping ei(α)to ei(β).

If α = (αi)ni=1 is a finite sequence, we define the finite dimensional space Xd(α) in the

same way.

To reduce ourselves to working with the spaces Xd(α) (and hence to sequences ofscalars), define, for a compact A ∈ B(H,K), the sequence α = D(A): if A has rankn < ∞, let α1 ≥ . . . ≥ αn > 0 be the non-zero singular values of A, and set αi = 0for i > n. If the rank of A is infinite, let (αi) be the singular values of A, listed in thenon-increasing order. We have:

Proposition 4.1 If H and K are separable infinite dimensional Hilbert spaces, andA ∈ B(H,K) is a compact contraction, then X(A) is 26 -completely isomorphic toXd(D(A)).



Proof Let α = (αi) = D(A). By Lemma 3.2(1), we can assume that A = |A| =(A∗A)1/2 , and H = K . If A has finite rank, (4–1) shows that X(A) is completelyisometric to Xd(α). Otherwise, write A =

�i αiξi ⊗ ξi , for some orthonormal system

(ξi)∞i=1 in H . Let P be the orthogonal projection onto K = span[ξi : i ∈ N]. SetQ = I − P, and L = Q(H). By Lemma 3.2(2), X(A) is

√2-completely isomorphic

to X(K,K,PAP)⊕X(L, L,QAQ). Furthermore, QAQ = 0, hence X(L, L,QAQ) = LR .It is easy to see that X(K,K,PAP) = Xd(α). By Proposition 3.7, the latter space is4√

2-completely isomorphic to Xd(α) ⊕ R. Thus,

X(A)√

2� X(K,K,PAP) ⊕ X(L, L,QAQ)

4√

2� Xd(α) ⊕ R ⊕ LR

√2

� Xd(α) ⊕ (�2 ⊕ L)R = Xd(α) ⊕ R 4√

2� Xd(α).

We will also use a related observation.

Lemma 4.2 For every compact contraction A ∈ B(�2), and every ε > 0, there existsα ∈ c0 such that X(A)

1+ε� Xd(α).

Proof Indeed, let β = D(A). If A is finite rank, then X(A) is completely isometricto Xd(β). If rank A = ∞, assume (by Lemma 3.2) that A = |A|. Then β1 ≥ β2 ≥

. . . > 0 is the list of all positive eigenvalues of A. Denote the corresponding norm 1eigenvectors by ξi . Let H = span[ξi : i ∈ N], and K = ker A. Clearly, K and H aremutually orthogonal subspaces of �2 . Let (ηj)j∈J be the orthogonal basis of K . Find asequence (γj)j∈J of positive numbers, satisfying

�j∈J γ

2i < ε2 . Consider the compact

contraction A ∈ B(�2), defined by Aξi = βiξi , and Aηj = γjηj . Then �A − A�2 < ε.By (4–1), the formal identity map id : X(A) → X(A) is a complete contraction, and�id−1�cb < 1 + ε. We complete the proof by identifying X(A) with the space Xd(α),where the sequence α = (αi) is the “join” of the sequences β and γ (that is, anynumber c ∈ [0, 1] occurs in α as many times as it occurs in the sequences β and γcombined).

Now denote by S the set of all sequences (αi)i∈N satisfying 1 ≥ α1 ≥ α2 ≥ . . . ≥ 0,and limi αi = 0. The rest of this section is devoted to the spaces Xd(α) (α ∈ S ).We translate the relations between sequences α,β ∈ S to relations between thecorresponding spaces Xd(α) and Xd(β). We say that the sequence α dominates β(α � β ) if there exists a set S ⊂ N) and K > 0 s.t.

�i∈S β

2i < ∞, and Kαi ≥ βi for

any i /∈ S . We say α is equivalent to β (α ∼ β ) if α � β , and β � α .

Clearly, the relation � is reflexive and transitive. The relation ∼ is, in addition to this,symmetric. For instance, to establish the transitivity of �, suppose α � β , and β � γ ,



and show that α � γ . Note that there exist sets S1 and S2 , and constants K1 and K2 ,s.t. βi � K1αi for i /∈ S1 , γi � K2βi for i /∈ S2 ,

�i∈S1

β2i < ∞, and

�i∈S2

γ2i < ∞.

Let S = S1 ∪ S2 , and K = K1K2 . Then γi � Kαi for i /∈ S . Moreover,�

i∈S

γ2i =

�

i∈S2

γ2i +

�

i∈S1\S2

γ2i �

�

i∈S2

γ2i +

�

i∈S1

K22β

2i < ∞,

which is what we need. The other properties are proved in a similar fashion.

Proposition 4.3 For α,β ∈ S , α ∼ β if and only if there exists a set S ⊂ N and aconstant K s.t.

�i∈S(α2

i + β2i ) < ∞, and K−1αi � βi � Kαi for i /∈ S .

Proof If S and K with the properties described above exist, then they witness the factthat α ≺ β and α � β , hence α ∼ β . Conversely, suppose α ≺ β and α � β . Thenthere exist a constant K , and sets S1 and S2 , s.t. αi � Kβi for i /∈ S1 , βi � Kαi

for i /∈ S2 ,�

i∈S1α2

i < ∞, and�

i∈S2β2

i < ∞. By reducing S1 and S2 further,we can assume that αi > Kβi for each i ∈ S1 , and βi > Kαi for each i ∈ S2 .Then

�i∈S1

β2i < ∞, and

�i∈S2

α2i < ∞. Therefore, S = S1 ∪ S2 has the required

properties.

The main result of this section is:

Theorem 4.4 For α,β ∈ S , Xd(α) embeds completely isomorphically into Xd(β) ifand only if α ≺ β .

From this we immediately obtain

Corollary 4.5 Suppose α,β ∈ S . The following three statements are equivalent.

(1) Xd(α) is completely isomorphic to Xd(β).

(2) Xd(α) embeds completely isomorphically into Xd(β), and vice versa.

(3) α ∼ β .

Proof (1) ⇒ (2) is trivial, while (2) ⇒ (3) follows from Theorem 4.4 and Proposi-tion 4.3. To establish (3) ⇒ (1), suppose α ∼ β . By Proposition 4.3, there exists aset I and K > 0 s.t.

�i∈I(α2

i + β2i ) < ∞, and K−1αi � βi � Kαi for any i /∈ I .

By Corollary 3.5, the spaces Eα = span[ei : i ∈ I] �→ Xd(α) and Eβ = span[ei : i ∈I] �→ Xd(β) are completely isomorphic to R. By (4–2), the formal identity map fromFα = span[ei : i /∈ I] �→ Xd(α) to Fβ = span[ei : i /∈ I] �→ Xd(β) is a completeisomorphism. By Lemma 3.2(2), Xd(α) � Eα⊕Fα , and Xd(β) � Eβ⊕Fβ . Therefore,the formal identity map from Xd(α) to Xd(β) is a complete isomorphism.



The proof of Theorem 4.4 follows from the next two lemmas.

Lemma 4.6 If α,β ∈ S and β ≺ α , then Xd(β) embeds completely isomorphicallyinto Xd(α).

Proof By Corollary 3.8, Xd(α) is completely isomorphic to X(A), where A ∈ B(�2⊕2�2) is defined by A = diag (α) ⊕ 0. Let (ξi) and (ξ�i) be orthonormal bases in thefirst and second copies of �2 , respectively. Then X(A) is the closed linear span of thevectors ξi⊕αiξi and ξ�i ⊕0 (i ∈ N). Find a sequence φi ∈ [0,π/2] s.t. βi = cosφi ·αi .Define an orthonormal system ηi = cosφiξi + sinφiξ�i . For i ∈ N consider

fi = ηi ⊕ βiξi = cosφi(ξi ⊕ αiξi) + sinφi(ξ�i ⊕ 0).

Then fi ∈ X(A), and span[fi : i ∈ N] = Xd(β).

Lemma 4.7 Suppose α,β ∈ S , E is a subspace of Xd(α), and a completely boundedmap U : E → Xd(β) has bounded inverse (in the terminology of [18], E is completelysemi-isomorphic to Xd(β)). Then β ≺ α .

Proof We rely heavily on Wielandt’s Minimax Theorem ([2, Theorem III.6.5]): ifc1 ≥ c2 ≥ . . . ≥ 0 are eigenvalues of positive compact operator T , then, for any finiteincreasing sequence i1 < . . . < ik of positive integers,

k�

j=1

cij = supE1�→...�→Ek

minxj∈Ej, (xj) orthonormal

�Txj, xj�,

where the supremum is taken over all subspaces E1 �→ . . . �→ Ek of the domain of T ,with dim Ej = ij for 1 � j � k . Actually, the theorem is stated in [2] for operators onfinite dimensional spaces, but a generalization to compact operators is easy to obtain.Applying the above identity to T = S∗S , where S is a compact operator with singularnumbers s1 ≥ s2 ≥ . . . ≥ 0, we obtain:

(4–3)k�

j=1

s2ij = sup

E1�→...�→Ek,dim Ej=ijmin

xj∈Ej, (xj) orthonormal�Sxj�

2.

In our situation, assume �U�cb = 1. Let c = �U−1�, A = diag (αi), B = diag (βi),B� = BU . Denote the singular values of B� by (β�

i ). Clearly, β�i � βi � cβ�

i for everyi ∈ N.



Pick an orthonormal system (xj)kj=1 in E (the domain of U ). Let u be the formal

identity from Rk = (�k2)R (the k-dimensional row space) to span[xj : 1 � j � k]. Then

(compare with the proof of Lemma 3.3)

�u�2cb = max{1, �Au�2

2} � 1 +k�

j=1

�Axj�2,

and (since U is a complete contraction)

�u�2cb ≥ �Uu�2

cb ≥ �BUu�22 =

k�

j=1

�B�xj�2.

Thus,�k

j=1 �B�xj�2 � �k

j=1 �Axj�2 +1 for any orthonormal family (xj)k

j=1 . By (4–3),

(4–4) 1 +k�

j=1

α2ij ≥

k�

j=1

β�2ij

for any i1 < . . . < ik (indeed, when computing�k

j=1 α2ij , we are taking the supremum

over a larger family of subspaces (Ej), than when we are computing�k

j=1 β�2ij ).

Now let I = {i ∈ N : β�i > 2αi}. Then

�i∈I β

�2i � 2. Indeed, otherwise there exists

a sequence i1 < . . . < ik of elements of I s.t. C =�k

j=1 β�2ij > 2. Then, by (4–4),

�kj=1 α

2ij ≥ C − 1. On the other hand,

�kj=1 α

2ij � C/2, a contradiction.

Developing the ideas of this proof, we obtain:

Theorem 4.8 Suppose α,β ∈ S , and there exists an isomorphism U : Xd(α) →

Xd(β) with �U�cb, �U−1�cb � C . Then the formal identity map id : Xd(α) → Xd(β)satisfies �id�cb, �id−1�cb < 4C2 .

Proof As in the proof of Lemma 4.7, let A = diag (αi), B = diag (βi), and B� = BU .By Lemma 3.3, �B�u�2 � C max{�Au�2, �u�} for any u : �2 → Xd(α). Denote thesingular numbers of B� by (β�

i ), and note that βi/C � β�i � Cβi for every i. Reasoning

as in the proof of Lemma 4.7, we see that

C2(1 +k�

j=1

α2ij) ≥

k�

j=1

β�2ij

Let I = {i : β�i > 2Cαi}. As in the preceding proof,

�i∈I β

�2i < 2C2 . Therefore,�

i∈I β2i < 2C4 , and βi � 2C2αi for i /∈ I .



Next we show that �id : Xd(α) → Xd(β)�cb < 4C2 . As before, denote the canonicalbases in Xd(α) and Xd(β) by (ei(α))i∈N and (ei(β))i∈N , respectively. By (4–2),Y1 = span[ei(α) : i ∈ I] and Y0 = span[ei(α) : i /∈ I] are completely contractivelycomplemented subspaces of Xd(α). Moreover,

�id|Y1�cb � �V : (Y1)R → Xd(β)�cb = max{1, �BV�cb}

� max {1,��

i∈Iβ2

i�1/2

} < 2C2

(here, V is the formal identity from (Y1)R to span[ei(β) : i ∈ I]). By Lemma 3.3 and(4–2), �id|Y0�cb < 2C2 . Therefore,

�id : Xd(α) → Xd(β)�cb � �id|Y1�cb + �id|Y0�cb < 4C2.

The norm of id : Xd(β) → Xd(α) is computed the same way.

5 Completely unconditional bases

In this section, we further investigate bases in spaces X(A), where A ∈ B(H,K) isa compact contraction. A subspace E of X(A) is isometric to X(A|E). As A|E isa compact contraction, (4–1) implies that E has a 1-completely unconditional basis.The key result of this section is Proposition 5.2, establishing the uniqueness of acompletely unconditional basis in E (the existence of such a basis has been establishedby Proposition 4.1). We also show that the canonical basis (and therefore, everycompletely unconditional basis) in a completely complemented subspace of Xd(α) isequivalent to a subsequence of the canonical basis of Xd(α). Moreover, there existsα ∈ S such that the canonical basis in every complemented subspace of Xd(α) isequivalent to a subsequence of the canonical basis of Xd(α) (Theorem 5.6). In general,the last statement need not be true (Remark 5.5).

First we show that any completely unconditional basic sequence in an X(A) spacecorresponds to a canonical basis of Xd(β), for some β .

Proposition 5.1 Suppose I is a finite or countable set, and (e�i)i∈I is a a C-completelyunconditional basic sequence in X(A) (A ∈ B(�2) is a contraction, not necessarilycompact). Let Y = span[e�i : i ∈ I], and define the sequence β = (βi) by settingβi = �Ae�i�. Consider the operator T : Y → Xd(β) : e�i �→ ei , where (ei) is thecanonical basis of Xd(β). Then �T�cb � C and �T−1�cb � C2 .



Proof As noted in Section 2,

(5–1) C−2�

�αie�i�

2 ��

|αi|2 � C2

�

�αie�i�

2

for any finite sequence of scalars (αi). Thus, �T�, �T−1� � C .

Let B = diag (β) (note that B = B∗ ). By Lemma 3.3, it suffices to show that

(5–2) �BTu�2 � C max{�Au�2, �u�}

for any u : �2 → Y = X(A|Y ), and

(5–3) �AT−1u�2 � C2 max{�Bu�2, �u�}

for any u : �2 → Xd(β). By Lemma 3.3, the complete unconditionality of (e�i) implies:

(5–4)�AΛu�2 � C max{�Au�2, �u�},�Au�2 � C max{�AΛu�2, �u�}

whenever Λ = diag (λi) (that is, Λe�i = λie�i ), with λi = ±1 for each i, and u ∈

B(�2, Y). Note that �Au�22 = tr(A∗Av), where v = uu∗ . Therefore, (5–4) is equivalent

to

(5–5)tr(Λ∗A∗AΛv) � C2 max{tr(A∗Av), 1},tr(A∗Av) � C2 max{tr(Λ∗A∗AΛv), 1}

whenever v ≥ 0 and �v� = 1. But �Λ∗A∗AΛe�i, e�j� = λiλj�Ae�i,Ae�j�. Averagingover λi = ±1 for each i, we see that AveΛΛ∗A∗AΛ = T∗B2T (since �T∗B2Te�i, e�j� =δij�Ae�i,Ae�j�, where δij is Kronecker’s delta). Therefore, by (5–5),

tr(T∗B2Tv) = AveΛtr(Λ∗A∗AΛv)� sup

Λtr(Λ∗A∗AΛv) � C2 max{tr(A∗Av), 1}

whenever v is a positive contraction. Thus, for any contraction u,

�BTu�22 = tr(T∗B2Tuu∗)� C2 max{tr(A∗Auu∗), 1} = C2 max{�Au�2, �u�}2,

which proves (5–2).

To establish (5–3), we show that, for every w : �2 → Y , we have

(5–6) �Aw�2 � C max{�BTw�2, �w�}

Indeed, let w = T−1u. Then, by (5–6) and (5–1),

�AT−1u�2 = �Aw�2 � C max{�BTw�2, �w�}

� C max{�Bu�2,C�u�} � C2 max{�Bu�2, �u�}.



Moreover, (5–6) is equivalent to the following: for any non-negative, norm one v ∈

B(Xd(α)), we have

(5–7) C2 max{tr(T∗B2Tv), 1} ≥ tr(A∗Av).

As we have established before, tr(T∗B2Tv) = AveΛtr(Λ∗A∗AΛv). By the linearity andpositivity of the trace,

0 � infΛ

tr(Λ∗A∗AΛv) � tr(T∗B2Tv) � supΛ

tr(Λ∗A∗AΛv).

Thus, for some Λ, tr(Λ∗A∗AΛv) � tr(T∗B2Tv). By (5–5),

tr(A∗Av) � C2 max{tr(Λ∗A∗AΛv), 1} � C2 max{tr(T∗B2Tv), 1},

which implies (5–7).

Proposition 5.2 For α ∈ c0 , the completely unconditional basis in Xd(α) is unique (upto permutative equivalence). More precisely: if (gi) is a C-completely unconditionalbasis in Xd(α), then it is 16C11 -equivalent (up to a permutation) to the canonical basisin Xd(α).

Proof Let (ei(α)) be the canonical basis of Xd(α). Set βi = �Agi�, and let (ei(β)) thecanonical basis of Xd(β). By Proposition 5.1, the map T : Xd(α) → Xd(β) : gi �→ ei(β)satisfies �T�cb � C , �T−1�cb � C2 . By Theorem 4.8, id : Xd(β) → Xd(α) satisfy�id�cb, �id−1�cb < 4C4 . Thus, the operator U = id ◦ T is a complete isomorphism onXd(α), with Ugi = ei , �U�cb < 4C5 , and �U−1�cb < 4C6 .

Corollary 5.3 Suppose α ∈ S , and Y is a C-completely complemented subspace ofXd(α). Then Y is 26C2 -completely isomorphic to Xd(α�), where α� is a subsequenceof α .

Proof Let P be a projection from Xd(α) onto Y , with �P�cb � C . Then Xd(α) is2C-completely isomorphic to Y ⊕ Z , where Z = ker P. By Lemma 4.2, Y and Zare

√2-completely isomorphic to Xd(β) and Xd(β�), respectively, where β and β�

belong to S . By Lemma 3.2(2), Xd(α) is 4C-completely isomorphic to Xd(γ), whereγ = (γi) ∈ S is the “join” of β = (βi) and β� = (β�

i ). More precisely, the sequence γhas the property that, for every c ∈ [0, 1],

|{i : γi = c}| = |{i : βi = c}|+ |{i : β�i = c}|.

Denoting the canonical basis of Xd(γ) by (ei(γ)), we see that Xd(β) = span[ei(γ) :i ∈ I], for some infinite set I ⊂ N. By Theorem 4.8, the formal identity id :Xd(γ) → Xd(α) : ei(γ) �→ ei(α) satisfies �id�cb, �id−1�cb < 8C . In particular, Xd(β)26C2 -completely isomorphic to span[ei(α) : i ∈ I].



Remark 5.4 By [17], the completely unconditional basis in R ⊕ C is unique up to apermutation.

Remark 5.5 In general, the canonical basis of a subspace of Xd(α) (α ∈ S ) need notbe equivalent to a subsequence of the canonical basis of Xd(α). For instance, supposethe sequence α = (αi) and β = (βi) are defined by setting αi = 2−n2 , βi = 2−n2−n

for 4n2 � i < 4(n+1)2 (n ∈ {0} ∪ N). By Lemma 4.6, Xd(β) embeds completelyisomorphically into Xd(α). However, (ei(β)) (the canonical basis of Xd(β)) is notequivalent to any subsequence of the canonical basis (ei(α)) of Xd(α). Indeed, suppose,for the sake of achieving a contradiction, that there exists a complete isomorphism Tfrom Xd(β) to a subspace of Xd(α), mapping ei(β) to eki(α). Fix n ∈ N withmax{�T�cb, �T−1�cb} < 2n/2 . Consider the sets

In = {4n2 � i < 4(n+1)2 : ki < 4(n+1)2},

Jn = {4n2 � i < 4(n+1)2 : ki ≥ 4(n+1)2}.

By Pigeon-Hole Principle, with In or Jn has the cardinality grater than 4n2+2n . If|In| > 4n2+2n , consider

x =�

i∈In

Ei1 ⊗ ei(β) ∈ M4(n+1)2 (Xd(β))

(recall that Ei1 is the “matrix unit” with 1 on the intersection of the first column andthe i-th row, and zeroes everywhere else). By (4–2),

�x�2M

4(n+1)2 (Xd(β)) = max{1, |In| · 2−n2−n} = |In| · 2−n2−n.

However, by (4–2) again,

�(IM4(n+1)2

⊗ T)x�2M

4(n+1)2 (Xd(α)) ≥ max{1, |In| · 2−n2−n} = |In| · 2−n2

,

yielding �T�2cb ≥ 2−n . If Jn > 4n2+2n , consider

x =�

i∈Jn

Ei1 ⊗ ei(β) ∈ M4(n+1)2 (Xd(β)).

As before, �x�2 = |Jn| · 2−n2−n , while

�(IM4(n+1)2

⊗ T)x�2M

4(n+1)2 (Xd(α)) ≥ max{1, |In| · 2−(n+1)2} = |Jn| · 2−(n+1)2

,

hence �T−1�2cb ≥ 2n+1 . Thus, max{�T�cb, �T−1�cb} ≥ 2n/2 , which yields the desired

contradiction.

In certain situations, the canonical basis for every subspace of Xd(α) is equivalent to asubsequence of the canonical basis of Xd(α).



Theorem 5.6 For any a > 1 there exists α ∈ S\�2 such that any subspace Y of Xd(α)(finite or infinite dimensional) has a 1-completely unconditional basis, a-equivalent toa subsequence of the canonical basis of Xd(α).

Combining this result with Proposition 5.2, we obtain

Corollary 5.7 There exists α ∈ S\�2 such that any C-completely unconditional basicsequence in Xd(α) is ACB -equivalent to a subsequence of the canonical basis of Xd(α)(here, A and B are positive).

Proof of Theorem 5.6 Assume a < 2. Pick a sequence of integers 1 = N0 < N1 <. . ., s.t. Nk > 2Nk−1 for each k . Define a sequence α = (αi) by setting α2i = a−k

for Nk � i < Nk+1 , α2i−1 = 0 for any i ∈ N. Clearly, α ∈ c0\�2 . Let A = diag (α).For a subspace Y of X(A), let β = (βi) is the sequence of singular values of A|Y .Define the set I1 by setting I1 = {1, . . . ,M} if rank (A|Y ) = M < ∞, and I1 = Nif rank (A|Y ) = ∞. We can also assume that the elements of (βi)i∈I1 are listed in thenon-increasing order. Then βi � α2i for each i.

Denote the normalized eigenvectors of (A|Y )∗A|Y , corresponding to the eigenvalues βi

(i ∈ I1 ), by ηi . Furthermore, find the vectors (ηi)i∈I0 , forming an orthonormal basisin ker (A|Y ). For i ∈ I0 , set βi = 0. Let I = I1 ∪ I0 (we assume that this union isdisjoint). Then the family (ηi)i∈I is the canonical basis for Y .

For each positive integer k , let Mk be the smallest value of i s.t. βi � a−k . SetM0 = 1. In this notation, a1−k ≥ βi > a−k iff Mk−1 � i < Mk . As noted above,βi � α2i , hence Mk � Nk . By our choice of the sequence (Nk),

Mk − Mk−1 < Mk � Nk � Nk+1 − Nk.

Thus, there exists an injective map π : I → N s.t. π(I0) ⊂ {2i − 1 : i ∈ N}, andπ([Mk−1,Mk)) ⊂ {2i : i ∈ [Nk,Nk+1)} for each k ∈ N. For i ∈ I0 , απ(i) = βi = 0,while for i ∈ I1 , aαπ(i) = a1−k ≥ βi > a−k = απ(i) . Define the operator T : Y →

span[eπ(i) : i ∈ I] �→ Xd(α), defined by Tξi = eπ(i) . By (4–1), T is a completecontraction, and �T−1�cb � a.

Remark 5.8 The proof of Theorem 5.6 shows that any subspace of Xd(α) is a-completely isomorphic to a completely contractively complemented subspace of Xd(α).Nevertheless, Xd(α) contains subspaces which are not completely complemented. Toconstruct them, find a sequence (βi) such that 1 ≥ β1 ≥ β2 ≥ . . . > 0, limβi = 0,and furthermore,

�i γ

2i = ∞, where γi = αiβi . Denote the canonical basis of Xd(α)



by (ei)i∈N . Then e2i−1 = ξ2i−1 ⊕ 0, and e2i = ξ2i ⊕αiξ2i ((ξj) is an orthonormal basisin �2 ). For k ≥ 0 and i ∈ [Nk, nk+1), let

(5–8) fi = βie2i +�

1 − β2i e2i−1 =

�βiξ2i +

�1 − β2

i ξ2i−1�⊕ γiξ2i.

We show that Y is not completely complemented in Xd(α). Indeed, suppose, forthe sake of contradiction, that there exists a c.b. projection P from Xd(α) onto Y .For ε ∈ {−1, 1}N , define an operator Λε ∈ B(Xd(α)) by setting λεej = ε�j/2�ej .For any such ε, (4–2) implies that Λε is a complete isometry. For any i ∈ N,we have Λεξ = εiξ whenever ξ ∈ span[e2i−1, e2i]. In particular, Λεfi = εifi . LetQ = Aveε∈{−1,1}NΛεPΛε . Note that ran Q ⊂ Y , and Q|Y = IY , hence Q is a projectiononto Y . Furthermore, �Q�cb � �P�cb .

For each i, we have Qe2i = aifi , and Qe2i−1 = bifi . The equations Qfi = fi and (5–8)yield aiβi + bi

�1 − β2

i = 1. Then supi max{|ai|, |bi|} � �Q�. As limβi = 0, thereexists K ∈ N such that |bi| > 1/2 for i > K . Find N ∈ N s.t.

�K+Ni=K+1 γ

2i > 4�Q�2

cb(this is possible, since

�i γ

2i = ∞). Consider x =

�K+Ni=K+1 Ei1 ⊗ e2i−1 ∈ MN(Xd(α)).

Then �x� = 1, and therefore,

�Q�cb ≥ �(IMN ⊗ Q)x� = �

K+N�

i=K+1

Ei1 ⊗ bifi�

≥ �

K+N�

i=K+1

Ei1 ⊗ biγiξ2i�MN (C) =� K+N�

i=K+1

|bi|2γ2

i

�1/2

≥

��K+Ni=K+1 γ

2i

�1/2

2> �Q�cb,

which is impossible.

Remark 5.9 If any subspace of a Banach space E is complemented, then E isisomorphic to a Hilbert space. This was first proved in [14], see also [10] for sharperasymptotics of the isomorphism constants. A recent preprint [9] exhibits a class ofseparable Banach spaces E , not isomorphic to �2 , such that every subspace of E isisomorphic to a complemented subspace of E . It is not known whether E can beconstructed in such a way that all of its subspaces have an unconditional basis.

6 Completely isomorphic classification of subspaces of X(A)

The main goal of this section is to prove Theorem 6.1 and Corollary 6.2 below. Recallthat C is the set of compact contractions which are not Hilbert-Schmidt.



Theorem 6.1 If A ∈ B(�2) belongs to C, then (S(X(A)),�) is Borel bireducible tothe complete Kσ relation.

Together with Corollary 4.5, this theorem immediately implies

Corollary 6.2 If A ∈ B(�2) belongs to C, then the relation of complete biembed-dability on S(X(A)) is Borel bireducible to the complete Kσ relation.

The proof of Theorem 6.1 proceeds in two steps. First, we introduce the space SA ofsequences of non-negative generalized integers, with an equivalence relation ∗

∼, andshow the latter is Borel bireducible with (S(X(A)),�). Then we prove that (SA,

∗∼) is,

in fact, a complete Kσ relation.

Suppose A ∈ B(�2) is of class C. By Lemma 3.2, we can assume that �A� < 1, andA ≥ 0. List the positive eigenvalues of A in the non-increasing order: 1 > �A� =so

1 ≥ so2 ≥ . . . > 0. In the terminology of Section 4, (so

i )i∈N = D(A). Clearly,limi so

i = 0. Let (ξi)i∈N be the normalized eigenvectors of A, corresponding to theeigenvalues so

i . We can identify span[ξi : i ∈ N] with �2 . Consider the operatorA = diag (so

i ) ⊕ 0 ∈ B(�2 ⊕ �2). By Corollary 3.8, X(A) � X(A). For the rest of thissection, we assume that A = A.

Denote by (ξ�i)i∈N the canonical orthonormal basis in the second copy of �2 . Then thecanonical basis of X(A) is the collection of vectors ei = ξi ⊕ so

i ξi and fi = ξ�i ⊕ 0. Asin (4–1), we have, for n × n matrices a1, b1, a2, b2, . . .,

(6–1)

�

�

i

ai ⊗ ei +�

i

bi ⊗ fi�2Mn(X(A))

= max {��

i

aia∗i +�

i

bib∗i �, ��

i

so2i a∗i ai�}.

For an infinite dimensional Y �→ X(A), we let (si(Y)) = D(A|Y ). Recalling thedefinition of D from Section 4, we see that, if rank (A|Y ) = ∞, then s1(Y) ≥ s2(Y) ≥. . . are the positive singular values of A|Y , listed in the non-increasing order. In thecase of rank (A|Y ) = n < ∞, s1(Y) ≥ s2(Y) ≥ . . . ≥ sn(Y) are the n positive singularvalues of A|Y , and si(Y) = 0 for i > n. In this notation, so

k = sk(X(A)). Clearly,sk(Y) � so

k for any k , and any Y �→ X(A).

For k ∈ N, set nk(Y) = sup{� ∈ N : 21−� ≥ sk(Y)}. The sequence n(Y) = (nk(Y))k∈Nbelongs to NN

∗ , where N∗ = N∪ {∞} is viewed as the 1-point compactification of N.For k ∈ N let αk = nk(X(A)). Define SA as the set of all elements β = (βi)i∈N ∈ NN

∗ ,



such that (1) βk ≥ αk for any k , and (2) β1 � β2 � . . .. Equipping NN∗ with its

product topology, we see that SA is closed.

For any infinite dimensional Y �→ X(A), the sequence (nk(Y))k∈N belongs to SA .Conversely, for any β ∈ SA there exists Y �→ X(A) s.t. β = n(Y). Indeed, supposeβk ∈ N∗ , βk ≥ αk for any k , and β1 � β2 � . . .. Let gi = sinφiei + cosφifi , withso

i sinφi = 2−βi . We denote span[gi : i ∈ N] by Y(β), where β = (βi). By (6–1),n(Y(β)) = β .

Define the relation ∗∼ on SA as follows: β

∗∼ γ if there exists K ∈ N and I ⊂ N s.t.

|βi − γi| � K for any i /∈ I , and�

i∈I(4−βi + 4−γi) � K . By Corollary 4.5, β ∗

∼ γ iffY(β) � Y(γ), and conversely, Y � Z iff n(Y) ∗

∼ n(Z).

Proposition 6.3 n and Y are Borel maps.

This immediately yields:

Corollary 6.4 (SA,∗∼) and (S(X(A)),�) are Borel bireducible to each other.

Proof of Proposition 6.3 First we handle the map Y. We have to show that, for anyopen set U ⊂ X(A), {β ∈ SA : Y(β) ∩ U �= ∅} is Borel. But Y(β) ∩ U �= ∅ iffthere exist m ∈ N and λ1, . . . ,λm ∈ Q + iQ s.t.

�mi=1 λigi ∈ U . Here, the vectors

(gi) come from the definition of Y. Note that, for each i, gi depends solely (andcontinuously) on βi . Therefore, for each m-tuple (λi)m

i=1 ,�m

i=1 λigi ∈ U is an opencondition on β . Thus, {β ∈ SA : Y(β) ∩ U �= ∅} is Borel.

Now consider n. Fix m,βm ∈ N, and show that the set of all Y ∈ S(X(A)), for whichnm(Y) > βm ∈ N, is Borel. To this end, find a countable set Om of orthonormalm-tuples ξ = (ξ1, . . . , ξm) in X(A), with the property that, for every ε > 0, and anyorthonormal m-tuple (η1, . . . , ηm) in X(A), there exists ξ = (ξ1, . . . , ξm) ∈ Om s.t.�ξi−ηi� < ε for any i. Furthermore, find a set Γm of m-tuples γ = (γ1, . . . , γm) ∈ Cm ,dense in the unit sphere of �m

2 .

The Minimax Principle (see e.g. [2, p. 75]) states that, for an operator T ∈ B(H,K),we have sm(T) > b iff H has an m-dimensional subspace E such that �Tη� > bfor any norm one η ∈ E . Therefore, nm(Y) > βm ∈ N (that is, sm(A|Y ) ≥ 2−βm )iff for every ε > 0 there exists an m-tuple of orthonormal vectors η1, . . . , ηm ∈ Y ,s.t. �

�mi=1 γiAηi� > 2−βm − ε whenever

�mi=1 |γi|

2 = 1. This, in turn, is equivalentto the following statement: for every r ∈ N, there exists (ξ1, . . . , ξm) ∈ Om s.t., for1 � i � m, Ba(ξi, 1/r) ∩ Y �= ∅ (here, Ba(x, c) denotes the open ball of radius c, withthe center at x), and �

�i γ

iAξi� > 2−βm −1/r for every (γi)mi=1 ∈ Γm . This condition

is Borel, hence n is Borel.



Lemma 6.5 (SA,∗∼) is a Kσ relation.

Proof We have to show that the set F = {(β, γ) ∈ SA × SA : β ∗∼ γ} is a Kσ set,

that is, a countable union of compact sets. To this end, define a family of subsets ofSA × SA , described below. For K, n ∈ N and In ⊂ {1, . . . , n}, define F(K, n, In) asthe set of all pairs (β, γ) (β = (βi), γ = (γi)) with the property that |βi − γi| � K fori ∈ {1, . . . , n}\In , and

�i∈In

(4−βi +4−γi) � K . Let F(K, n) = ∪In⊂{1,...,n}F(K, n, In),and F(K) = ∩nF(K, n). It suffices to show that F = ∪K∈NF(K). Indeed, F(K, n, In)is a compact subset of SA × SA , hence so is F(K, n) (as a finite union of compact sets).Furthermore, F(K) is also compact, and ∪KF(K) is Kσ .

We show first that F ⊂ ∪KF(K). By definition, β ∗∼ γ if there exists K ∈ N and

I ⊂ N s.t. |βi − γi| � K for any i /∈ I , and�

i∈I(4−βi + 4−γi) � K . Letting

In = I ∩ {1, . . . , n}, we see that (β, γ) ∈ F(K, n, In) for each n, hence (β, γ) ∈ F(K).

To prove the converse implication, suppose (β, γ) ∈ F(K) for some K , and show thatβ

∗∼ γ . Construct a tree T ⊂ {0, 1}N : for each n, T ∩ {0, 1}n consists of all the sets

In s.t. |βi − γi| � K for i /∈ In , and�

i∈In(4−βi + 4−γi) � K (we identify the set

of subsets of {1, . . . , n} with {0, 1}n ). The set T is indeed a tree: if In ∈ T , thenIn ∩ {1 . . . ,m} ∈ T for m < n. By assumption, T has arbitrarily long branches. ByKonig’s Lemma [11, p. 20], T has an infinite branch, which yields a set I ⊂ N s.t.|βi − γi| � K for i /∈ I , and

�i∈I(4

−βi + 4−γi) � K .

Next consider a space Ξ =�

k∈N Ξk , where Ξk = {0, . . . , k−1}, with the equivalencerelation cEKσb iff supi |ci −bi| < ∞ (here, c = (ci)i∈N , b = (bi)i∈N ). By [24], EKσ isa complete Kσ relation, hence, by Lemma 6.5, it reduces (SA,

∗∼). It remains to prove

the converse.

Proposition 6.6 There exists a Borel map φ : Ξ → SA s.t. φ(b) ∗∼ φ(c) iff cEKσb.

Proof Recall that A is a compact contraction which is not Hilbert-Schmidt. Therefore,�i 4−αi = ∞, and limi αi = ∞. Thus, there exists a sequence of positive integers

1 = p0 < q1 < p2 < q2 < . . . s.t.�

i∈Ik4−αi > 42k (Ik = [pk, qk − 1)), and

αpk+1 > k + qk . Define φ((bi)) = (b�j) by setting b�j = αj + bk if j ∈ Ik , andb�j = min{αj + k,αpk+1} if qk � j < pk+1 . Clearly, φ is a Borel map. Moreover, ifcEKσb, then φ(b) ∗

∼ φ(c). Suppose, on the contrary, that b� ∗∼ c� , where b� = φ(b) and

c� = φ(c). Then there exists a set I ⊂ N and K ∈ N s.t. |b�j − c�j| � K for j /∈ I , and�

j∈I(4−b�j + 4−c�j ) � K . We shall show that |bk − ck| � K for all but finitely many



k’s. Indeed, otherwise there exist infinitely many k’s s.t. Ik ⊂ I (this follows from thefact that b�j − c�j = bk − ck for j ∈ Ik ). But

�

j∈Ik

(4−b�j + 4−c�j ) ≥ 2 · 4−k�

j∈Ik

4−α�j > 1,

hence �

j∈I

(4−b�j + 4−c�j ) ≥�

Ik⊂I

�

j∈Ik

(4−b�j + 4−c�j ) = ∞,

a contradiction.

Conclusion of the proof of Theorem 6.1 By Corollary 6.4, (S(X(A)),�) and (SA,∗∼)

are Borel bireducible to each other. By Lemmas 6.5 and 6.6, (SA,∗∼) is Borel bireducible

to a complete Kσ relation.

Remark 6.7 For many separable Banach spaces X , it is known that the isomorphismrelation on S(X) reduces certain “classical” relations, such as EKσ (see e.g. [1, 4, 5, 6]).

7 Proofs of Theorems 1.1, 1.2, and 1.3

Recall that the class C consists of all compact contractions, which are not Hilbert-Schmidt, and the family F is the set of all operator spaces X(A), where A ∈ B(�2)belongs to C. Clearly, all these spaces are isometric to �2 .

Proof of Theorem 1.1 Suppose X(A) ∈ F. By Theorem 6.1 and Corollary 6.2,the relations of complete isomorphism and complete biembeddability on S(X(A)) arecomplete Kσ . To show that F contains a continuum of spaces, not completely iso-morphic to each other, pick A ∈ B(�2) ∩ C. Consider a space Ξ =

�k∈N Ξk , where

Ξk = {0, . . . , k − 1}, with the equivalence relation cEKσb iff supi |ci − bi| < ∞

(here, c = (ci)i∈N , b = (bi)i∈N ). By the results of Section 6, there exists a Borelmap Φ : Ξ → S(X(A)), such that Φ(b) � Φ(c) iff bEKσc. It remains to find a family(bε)ε∈{0,1}N ⊂ Ξ, such that bεEKσbδ iff ε = δ . To this end, write N as a disjoint unionof infinite sets Ik (k ∈ N). For any ε = (εk)∞k=1 , define

bε(i) =�

0 i ∈ Ik, ε(k) = 0i − 1 i ∈ Ik, ε(k) = 1

.

Clearly, this family (bε) has the desired properties.



Proof of Theorem 1.2 Consider A ∈ B(�2) of class C. The existence of the canonicalbasis has been established at the beginning of Section 4, while its uniqueness followsfrom Proposition 5.2.

Proof of Theorem 1.3 Combine Theorem 1.2 with Theorem 5.6 and Corollary 5.7.

8 Isometric classification: proof of Theorem 1.4

We handle the real case. Begin by introducing a numerical invariant of subspaces ofX = R ⊕1 �2 . Denote by P the “natural” projection onto R. For Y ∈ S(X), definec(Y) = �P|Y�.

Lemma 8.1 For Y ∈ S(X), there exists x ∈ Y such that �x� = 1, and �Px� = c(Y).Moreover, if this x is written as x = c(Y) ⊕ (1 − c(Y))ξ0 , then Y = span[x, Y �], whereY � = {0 ⊕ ξ : ξ ∈ (Y ∩ �2) ∩ ξ⊥0 }.

Proof If c(Y) = 0, the statement is trivial. Suppose c(Y) = 1. Then, for every n ∈ N,there exists tn ∈ (1 − 1/n, 1] and ξn ∈ �2 s.t. �ξn� = 1, and tn ⊕ (1 − tn)ξn ∈ X . AsY is closed, 1 ⊕ 0 ∈ Y .

Next consider c(Y) ∈ (0, 1). Suppose, for the sake of contradiction, that there is no x asin the statement of the lemma. Then for every n ∈ N there exist tn ∈ (c(Y)−1/n, c(Y)),and ξn ∈ �2 s.t. �ξn� = 1, and tn ⊕ (1 − tn)ξn ∈ Y . Passing to a subsequence ifnecessary, we can assume that (ξn) is a Cauchy sequence in �2 . Indeed, otherwisethere exist n1 < n2 < . . . and α > 0, such that, for any i, �ξni+1 − ξni� > α .By the uniform convexity of Hilbert spaces (which follows, for instance, from theparallelogram identity), there exists β > 0 s.t. �(1 − tni+1 )ξni+1 + (1 − tni)ξni�/2 <1 − c(Y) − β for any i. Define

yi =tni+1 + tni

2⊕

(1 − tni+1 )ξni+1 + (1 − tni)ξni

2∈ Y.

Then �yi� < 1 − β , and limi �Pyi� = c(Y). Therefore, �P|Y� > c(Y), which isimpossible.

Thus, the sequence (ξn) converges to some ξ0 ∈ �2 . Then x = c(Y) ⊕ (1 − c(Y))ξ0 isthe limit of the sequence tn ⊕ (1− tn)ξn , hence it belongs to Y . Clearly, �Px� = c(Y),and Y = span[x, Y ∩ �2]. Moreover, any ξ ∈ Y ∩ �2 is orthogonal to ξ0 . Indeed,



otherwise there exists ξ ∈ Y ∩ �2 and z ∈ C s.t. �ξ0 + zξ� < �ξ0�. Then x� =x + (0 ⊕ ξ) = c(Y) ⊕ (ξ0 + zξ) belongs to Y , �x�� < 1, and �Px�� = c(Y), which isimpossible.

For t ∈ [0, 1], define φ(t) = t +�

(1 − t)2 + 1. Clearly, φ is continuous andincreasing.

Lemma 8.2 For Y ∈ S(X),

φ(c(Y)) = sup { lim infi

�x + yi� : x, yi ∈ Y, �x� = �yi� = 1, yiw→ 0}.

Moreover, there exist a norm 1 x ∈ Y , and a normalized weakly null sequence (yi) inY , such that φ(c(Y)) = �x + yi� for every i.

Proof Assume c(Y) ∈ (0, 1) (only minimal changes are needed to handle c(Y) ∈

{0, 1}). Write x = t⊕ (1− t)ξ and yi = ti ⊕ (1− ti)ξi . Here, t, ti ∈ [0, 1], ξi ∈ �2 , and�ξi� = 1. As yi → 0 weakly, ti → 0, and �ξ, ξi� → 0. Therefore, limi �x+yi� = φ(ti).Taking the supremum over all x ∈ Y , we prove the desired equality. Furthermore, byLemma 8.1, Y = span[x, Y ∩ �2], where x = c(Y) ⊕ (1 − c(Y))ξ0 , ξ0 ∈ �2 has norm1, and Y ∩ �2 is orthogonal to ξ0 . Let (ξi) be an orthonormal basis in Y ∩ �2 . Thenφ(c(Y)) = �x + yi� for every i, and yi

w→ 0.

Lemma 8.3 If Y and Z are infinite dimensional subspaces of X , and Y is almostisometrically embeddable into Z , then c(Y) � c(Z).

Proof By definition, for every λ ∈ (1, 1.1), there exist a subspace W �→ Z and acontraction T : Y → W with �T−1� < λ. It suffices to show that

(8–1) φ(c(W)) ≥ λ−1φ(c(Y)) − 2(λ− 1).

Indeed, then we would conclude

φ(c(Z)) ≥ φ(c(W)) ≥ λ−1φ(c(Y)) − 2(λ− 1).

As the above inequality holds for any λ > 1, we conclude that φ(c(Y)) � φ(c(Z)). Bythe monotonicity of φ, c(Y) � c(Z).

By Lemma 8.2, there exists a normalized weakly null sequence (yi) in Y , and a normone x ∈ Y , such that φ(c(Y)) = �x + yi�. In the space W , consider the elementsx� = Tx/�Tx�, and y�i = Tyi/�Tyi�. Then the sequence (y�i) is weakly null, and

(8–2) �x� + y�i� ≥ �T(x + yi)� − �(1 − �Tx�−1)Tx� − �(1 − �Tyi�−1)Tyi�.



But �T−1��T(x + yi)� ≥ �x + yi�, hence �T(x + yi)� > λ−1φ(c(Y)). Furthermore,1 ≥ �Tx� > λ−1 , hence �(1−�Tx�−1)Tx� < λ− 1. Similarly, �(1−�Tyi�

−1)Tyi� <λ− 1. By (8–2), �x�+ y�i� > λ−1φ(c(Y))− 2(λ− 1). Applying Lemma 8.2, we obtain(8–1).

Lemma 8.4 If Y and Z are infinite dimensional subspaces of X , and c(Y) = c(Z),then Y is isometric to Z .

Proof We consider the case c(Y) = c(Z) ∈ (0, 1) (the extreme cases of c(Y) = c(Z) ∈{0, 1} are handled similarly). By Lemma 8.1, Y contains a norm one y = c(Y)⊕ξY ∈ Y(note that �ξY� = 1 − c(Y)), s.t. Y = span[y, Y ∩ �2], and Y � = Y ∩ �2 is orthogonalto ξY . Thus, for any ξ ∈ Y � , �y + ξ� = c(Y) +

�(1 − c(Y))2 + �ξ�2 . Similarly,

Z = span[z, Z�], and �z + η� = c(Z) +�

(1 − c(Z))2 + �η�2 for any η ∈ Z� . As Y �

and Z� are both infinite dimensional separable Hilbert spaces, there exists an isometryT � from Y � onto Z� . We complete the proof by defining the isometry T from Y ontoZ by setting Ty = z, and T|Y� = T � .

Proof of Theorem 1.4 By Lemmas 8.3 and 8.4, the following statements are equiv-alent for Y, Z ∈ S(X): (i) Y and Z are isometric, (ii) d(Y, Z) = 1, (iii) Y and Z areisometrically bi-embeddable, (iv) Y and Z are almost isometrically bi-embeddable, (v)c(Y) = c(Z). Denote that canonical basis for �2 by e0, e1, . . ., and consider a map

Φ : [0, 1] → S(X) : t → span[t ⊕ (1 − t)e0, 0 ⊕ e1, 0 ⊕ e2, . . .].

Then c(Φ(t)) = t , hence Φ(t1) and Φ(t2) satisfy any (equivalently, all) of the relations(i) – (iv) iff t1 = t2 . It remains to prove that the maps Φ and c are Borel.

To handle Φ, consider an open ball U ⊂ X with the center at α⊕�N

i=0 βiei and radiusr . Then Φ(t) ∩ U �= ∅ iff there exist λ0, . . . ,λN ∈ Q s.t.

|α− tλ0|+�|β0 − (1 − t)λ0|

2 +N�

i=1

|βi − λi|2�1/2

< r.

This inequality describes a Borel subset of [0, 1]. As any open subset of X is acountable union of open balls, the map Φ is Borel.

To deal with c, consider the sets

Ut = {s ⊕ ξ ∈ R⊕1 �2 : |s| > t, �ξ� <�

1 − t2}

(t ∈ [0, 1]). Clearly, Ut is an open subset of X , and c(Y) > t iff Y ∩ Ut �= ∅.

Remark 8.5 Theorem 1.4 holds not only for R ⊕1 �2 , but also for R ⊕p �2 , for1 � p < 2.



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Timur Oikhberg, Department of Mathematics, The University of California at Irvine, Irvine CA92697, and Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana,IL 61801Christian Rosendal, Department of Mathematics, Statistics, and Computer Science (M/C 249),University of Illinois at Chicago Chicago, IL 60607-7045

[email protected], [email protected]

Received: 24 September 2010 Revised: 1 March 2011


http://dx.doi.org/10.1007/s00222-002-0235-x

http://dx.doi.org/10.2178/jsl/1129642127

http://dx.doi.org/10.1007/s00208-005-0732-5

mailto:[email protected]

mailto:[email protected]

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